# Is there any relationship between frequency of signal and distance it travels?

I'm currently researching 5G technology. Many videos/articles on this topic claim that 5G is designed for communicating on lower distances (than 4G), because of higher (than 4G) frequency. So to build a network we need more dense grid of access points. Is this some general rule, that the higher frequency means less range, or is it connected to this exact frequency range, specific to 5G?

• Atmospheric absorption peaks at 60---70Ghz. – analogsystemsrf May 22 at 18:11
• Diffraction increases with frequency. FM has shorter range than AM because it uses higher carrier freq. Long waves can go by obstacles better. But increasing carrier wave increases BW for the information signal. – user1999 May 22 at 18:48
• @analogsystemsrf It's more complicated than that. 60 GHz has an oxygen absorption line, but then absorption goes back down until a second line at 118 GHz, then a water line at 183 GHz, etc. Water and other atmospheric gases have specific absorption regions and transmission windows all the way up through the infrared, where it clears up by the visible. – user71659 May 23 at 3:22

In general, yes, higher frequencies attenuate more the further distance they travel. There are two effects that are responsible for this.

First, higher frequency radio waves tend to be absorbed more readily by objects (ie: the penetration depth in the material is shorter). You may have noticed this effect with your home WiFi network. If you have a dual-band router, you probably have two networks, one at 2.4 GHz and one at 5 GHz. You may have noticed that the 2.4 GHz network reaches some corners of your house that the 5 GHz network does not - this is because the 2.4 GHz network penetrates walls better than the 5 GHz network does.

In terms of fifth-generation cell networks, this is the principal effect that limits range: people prefer to live in houses and apartment buildings, and those houses absorb more RF energy at higher frequencies.

The second effect is less intuitive, and falls out the Friis equation for free-space path loss.

$$FSPL= 20\log_{10}(d) + 20\log_{10}(f) + 20\log_{10}\left(\frac{4\pi}{c}\right)$$

Where $$\FSPL\$$ is the free space path loss in dB, $$\d\$$ is the distance in meters, and $$\f\$$ is the frequency in hertz, and $$\c\$$ is the speed of light. Note that as $$\f\$$ increases, the loss increases: a doubling in frequency results in a 6 dB increase in path loss.

This equation would lead you to believe that free space in some way attenuates higher frequencies more than lower frequencies, but that's not really the truth, even though it's a convenient lie we engineers tell ourselves. The "free space path loss" is actually a geometric effect that captures the "spread" of the electromagnetic wave as it propagates in free space, the same as how the beam of a flashlight spreads out when it's shone out into a big dark room.

The reason it's frequency dependent is because it does not account for the gain of the transmitting and receiving antenna - they are assumed to be isotropic. An isotropic antenna at a higher frequency is physically smaller than one at a lower frequency, which means that geometrically they have a smaller effective aperture.

If you keep the effective aperture of both antennas at both ends of the link the same physical size as you go up in frequency, you actually see that your path loss is independent of frequency. However, you would also find that the antennas are now directional: as you increase the aperture size, the transmitting antenna shoots a tighter and tighter beam, and the receiving antenna can only receive signals from a narrower and narrower cone. Since engineers usually think in terms of preserving a given antenna pattern and not in aperture size (since pointing antennas can be problematic), we commonly consider free space to have "loss" that increases with frequency when in reality that isn't quite the case.

• Doesn't look like inline math formatting with single enclosing dollar signs is working - anyone have any ideas about what I did wrong? – Peter May 22 at 18:29
• You need to use \$ instead of $ here. This is done differently on different sites due to the tradeoff of misformatting currency or code. – Kevin Reid May 22 at 18:33
• Thanks, that worked. – Peter May 22 at 18:35
• Thank you very much! It's really detailed answer. – Piotrek May 22 at 19:28